$\newcommand{\cl}{\operatorname{cl}}$Your $\Rightarrow$ argument is fine, save that it would be clearer if you said explicitly that $\overline A$ denotes closure in $X$. (This is why I prefer the notation $\cl A$, since it is so easily modified to show the space in which the closure is being taken: $\cl_XA$.)
You $\Leftarrow$ argument isn’t right: $X$ is a subspace containing $\cl A\cup\cl B$, so you’re not actually using the hereditary normality of $X$ at all. Let $Y=X\setminus(\cl_XA\cap\cl_XB)$; $\cl_YA$ and $\cl_YB$ are disjoint closed subsets of $Y$, and $Y$ is both normal and an open subset of $X$, so ... ?
Here is an example (simpler than the one I mentioned in the comments) of a space which is completely regular but not normal.
The Sorgenfrey line is the space $\Bbb R_\ell=(\Bbb R,\tau)$ where a basis for $\tau$ is given by $\{[a,b)\mid a,b\in\Bbb R\}$, this is a completely regular space (actually it is completely normal as well).
The Sorgenfrey plane is the space $\Bbb R_s=\Bbb R_\ell\times\Bbb R_\ell$ with the product topology, or in other words $\Bbb R^2$ where a basis of the topology is given by the half open squares. This space is still completely regular, as the product of two completely regular spaces is completely regular (easy exercise).
Note that the set $\Delta=\{(x,-x\}\mid x\in\Bbb R\}$ is an uncountable closed discrete subspace of $\Bbb R_s$ (draw a picture to convince yourself of this fact) and that this space is separable, since $\Bbb Q^2$ is a dense subset (every open set contains a Euclidean open set, so every open set intersects $\Bbb Q^2$).
Now if $\Bbb R_s$ were normal, then we could use Tietze's extension theorem to show that there are at least $2^{2^{\aleph_0}}$ continuous functions $\Bbb R_s\to\Bbb R$, since every function $\Delta\to\Bbb R$ is continuous and there are $2^{2^{\aleph_0}}$ distinct such functions, all of which could be extended to a function $\Bbb R_s\to\Bbb R$.
But $\Bbb R_s$ is separable, so a continuous function is determined by its values on $\Bbb Q^2$ and there's only $2^{\aleph_0}$ functions $\Bbb Q^2\to\Bbb R$, contradicting the previous paragraph. Note that this argument really shows that a separable space with an uncountable closed discrete subset cannot be normal (so you can also use it to show that the Moore plane, another example of a space which is completely regular but not normal, is in fact not normal).
Let me finish by mentioning two useful resources for this kind of questions: $\pi$-base, an online database of topological spaces, searchable by properties, and the book "counterexamples in topology" by Steen and Seebach, where $\Bbb R_\ell$ and $\Bbb R_s$ are the spaces 51 and 84 respectively.
Best Answer
A subspace $Y$ of a completely normal space $X$ is again completely normal: if $Z$ is a subspace of $Y$, then $Z$ is also a subspace of $X$ and so $Z$ is normal.
Another name for this property is "hereditarily normal". In general we can define, when we have a topological property $P$, the (probably) stronger property "hereditarily P": all subspaces of $X$ have property $P$.
For $P$ = "normal" this indeed is a stronger property than $P$. Hereditarily Lindelöf spaces are also a standard property, as are hereditarily separable spaces. In all these cases there are spaces with property $P$ that are not hereditarily $P$.
The process (if it can be called that) stops here though: a space that is hereditarily $P$ is also "hereditarily hereditarily $P$", using the same remark that a subspace of a subspace is still a subspace of the original space (transitivity of subspace topologies).